More calculatory bliss.
On a forum it was recently asked just how powerful the Death Star explosion over Yavin IV could've possibly been . . .that is, "what is the most energy released by the Death Star that Yavin IV could have survived?"
Well, that sounded like fun.
A beginning estimate would be to basically figure out a level of energy
that would observably damage the moon on the surface. This is useful
because of three factors.
1. As the Rebels take off to go fight
the Death Star, the red gas giant is visible in the sky. This suggests
that the Yavin base ought to have been within line of sight of the Death
2. After the battle, we get two useful scenes.
A. We see a brief shot of the planet from a great distance as the surviving Rebels fly home.
B. We see a close shot of the jungle near the Rebel base with its impenetrably dense stone before the medal ceremony.
as a result, we know that the condition of the base and nearby jungle
was basically the same before and after the Death Star was destroyed.
With that fact in mind, we can estimate an energy intensity that would
have produced observable effects, then rock the inverse square law back
out to the range of the Death Star. Since the planet and jungle are
not observably affected, this would serve as a back-of-the-envelope
Now, in reality, there would be
a number of other factors to consider, such as atmospheric attenuation
(especially in the sense of what parts of the spectrum make it through),
cloud cover more specifically, and so on. So while we could use
surface effects and ignore the atmosphere, technically the actual energy
would need to be greater. For sunlight on Earth, it is said that
about 50 percent makes it to the surface to be absorbed, 30 percent gets
reflected at the surface, and 20 percent is absorbed by the atmosphere.
That 20 percent is what we need to zoom in on at the moment, because I
presume the 30 percent figure is a general value including such things
as oceans (what with their being 75 percent of the planet and all),
which are not relevant to our inquiry.
Just in case, though,
let's assume that a third of the radiative emission of an exploding
Death Star would be absorbed by the atmosphere . . . that means we need
to increase our relevant values below by 33 percent.
for the range to the Death Star from Yavin, most estimates are in the
150,000km range, which I believe is true for both the SFJ pages and
Wong's SDN. (Also, some nincompoop on SDN claimed I am the source for a
very specific figure from Michael January's pages, but that's just a
funny thing I've been wanting to share and is not relevant right now.)
haven't personally studied the matter, but we'll roll with this figure
for now as it is not really inconsistent with the ~77,000km range of the
- Damaging the Surface
For ease, we can use estimates from http://glasstone.blogspot.com/2006/04/ignition-of-fires-by-thermal-radiation.html
. . . that for a nuclear explosion, the energy required to ignite
assorted things like dry and dark rotted wood falls somewhere in the
range of 15-25 calories per square centimeter (my own selected range
from the wide assortment of figures he reports). Now, Yavin IV didn't
exactly look dry, so we could crank this up further to 50 just to try to
be sure, and to try to cover the 33% atmosphere absorption mentioned
I don't see where a precise time factor is provided, but
we can presume for the purpose of this that the nuclear weapon and the
Death Star blast would radiate for about the same amount of time.
to joules per square centimeter, then, with a calorie being 4.184
joules, that's 209 J, which in square meters would be 2,090,000 . . . so, two megajoules per square meter. (The total energy incident on Yavin
would be interesting but not relevant here.)
- The Inverse Square Law
inverse square law is a fancy term for the idea that the intensity of
something per unit area at a range of, say, one meter (e.g. an intensity
of 1 watt per square centimeter at one meter's range) will fall off in a
geometric fashion. So, at two meters, you'll actually see the
intensity drop to 1/4th watt per square centimeter, because while the
distance has only doubled, the affected area will quadruple. At three
meters it's 1/9th, et cetera. And note that while I used watts above,
joules work just as well if you don't need a time factor . . . indeed,
the inverse square law applies to all sorts of things, what with it
being fairly basic geometry, applied.
The fun part for our
purposes, though, is that you can also estimate the strength of the
actual source itself if you know this intensity at a given range. This isn't really the inverse square law per se, but it is often discussed in the same breath.
the case of 1 watt per square centimeter at one meter's range, we can
work backward. It is 100 centimeters to the point where the intensity
is 1 watt per square centimeter, meaning that if we want to work
backwards to a 'zero point' all we need to do is get the total wattage
for the whole sphere. That's just figuring the whole area of the sphere
and multiplying the known wattage on one square centimeter accordingly.
So, a 100cm sphere has an area of 4πr^2, or 4π(100cm)^2, or
125,663.7cm^2. Since we know the intensity is 1 watt per square
centimeter and we know the whole sphere is 125,663.7 square centimeters,
then the point source of energy must be outputting at a rate of
125,663.7 watts, or about 126 kilowatts. (That's a lot, but so is a
watt per square centimeter at one meter's range. Per the basic
geometry, you'd be looking at 10,000 watts per square centimeter at 1
centimeter range . . . yet the sun only hits us with 0.14 watts per
square centimeter, and we only receive about 0.1 watts per square
centimeter at the surface.)
- Rocking the Law
150,000,000 meters as the range and two megajoules per square meter as
the energy intensity at that range, the total energy of the Death Star
blast would be 4π(150,000,000)^2 times two megajoules, or
282,743,338,823,081,391.5m^2 times two megajoules. That gives us about
565,500,000,000,000,000 megajoules, or 565,500 exajoules, or 5.655E23 . .
. or, about 135,000 gigatons.
If, however, the estimated range is wrong and it was really, say, 500,000 kilometers (500,000,000 meters), then the final value would be much higher, in the range of 6.3E24 J, or about 1.5 million gigatons.
The sun puts out 4E26J every
second, meaning that the Death Star explosion upper limit from the first example would be equivalent to 1/700th the energy output of the sun at the
range and with the energy on the surface specified.
Given a range of 500,000km, the comparison comes out rather better, with the Death Star explosion coming in at about 1/64th of the sun's energy every second.
However, given the way the Death Star II played out so very much closer to the surface of Endor, all of these figures seem ridiculously high.
seemed to recall doing something like this before, so I went to look
and found that this is a different technique (insofar as calculating via
affecting foliage) than what I used in regards to getting an upper
limit for the DS2 explosion in the Superlaser Effect section of my site.
See #4 at http://www.st-v-sw.net/STSWsuperl-5.html
and you'll see a technique involving evenly heating the atmosphere by
fifty degrees Celsius. While I still don't think that was a completely terrible approach, I think
the full methodology above is much superior.
Applying it, and using the 3500km value, we get 3E20 J as the upper limit, which is 1/1,333,333 of the sun's output in one second.
Even that is very much too high, given that we know exposed skin was unaffected by the DS2 blast . . . two megajoules per square meter in a short timespan like that would definitely provide a touch of sunburn. Leia didn't even seem to notice the DS2 had gone up until several seconds later, so we can reasonably suppose that she was not burned, blinded, et cetera.
3E20 joules . . . about 73.6 gigatons . . . is thus, to my mind, a very solid upper limit value for the DS2 destruction, and probably applies with caveats to the first Death Star, to boot.
Also, just for bonus points, and to actually apply the main part of the inverse square law a little, if we assume the Falcon and friends were only about 500 kilometers from the second Death Star when it blew, then they would've been getting about fifty times more energy per square meter, or 100 megajoules per square meter. The Falcon would therefore probably have had to deal with, say, 20,000 megajoules . . . 20 gigajoules . . . incident upon her, at worst case. That's a bit less than a five ton bomb.
Wow! I am really dumb because I still am not sure what your answer is. Would Yavin-IV/Endor have survived the DS's explosion?
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